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# Trig

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I need help with a trig problem, please.

Simplify.

(csc(x)-cot(x))*(sec(x)+1)

Guest Mar 12, 2017
edited by Guest  Mar 12, 2017

#1
+12565
+5

(1/sin - cos/sin)(1/cos +1)

(1-cos)/sin (1/cos +1)

(1-cos)/cos sin + (1-cos)/sin

(1-cos)/cos sin + (1-cos)(cos)/cos sin

{(1-cos) + (cos- cos^2) ] /  cos sin

(1-cos^2)/(cos sin)

sin^2 /(cos sin

sin/cos

tan(x)

ElectricPavlov  Mar 12, 2017
#1
+12565
+5

(1/sin - cos/sin)(1/cos +1)

(1-cos)/sin (1/cos +1)

(1-cos)/cos sin + (1-cos)/sin

(1-cos)/cos sin + (1-cos)(cos)/cos sin

{(1-cos) + (cos- cos^2) ] /  cos sin

(1-cos^2)/(cos sin)

sin^2 /(cos sin

sin/cos

tan(x)

ElectricPavlov  Mar 12, 2017
#2
+7155
+6

$$(\csc (x) - \cot (x))(\sec (x)+1) \\~\\ = (\frac{1}{\sin (x)}-\frac{\cos (x)}{\sin (x)})(\frac{1}{\cos (x)}+1) \\~\\ = (\frac{1-\cos (x)}{\sin (x)})(\frac{1}{\cos (x)}+\frac{\cos (x)}{\cos (x)}) \\~\\ = (\frac{1-\cos (x)}{\sin (x)})(\frac{1+\cos (x)}{\cos (x)}) \\~\\ = \frac{1 + \cos (x) - \cos (x) - \cos^2 (x)}{\sin (x) \cos (x)} \\~\\ = \frac{1 - \cos^2 (x)}{\sin (x) \cos (x)} \\~\\ = \frac{\sin^2 (x)}{\sin (x) \cos (x)} \\~\\ = \frac{\sin (x)}{ \cos (x)} \\~\\ = \tan (x)$$

hectictar  Mar 12, 2017